Use a substitution of the form \( u=a x+b \) to evaluate the following indefinite integral. \( \int \sqrt{2 x+5} d x \) \( \int \sqrt{2 x+5} d x=\square \)

To evaluate the integral \( \int \sqrt{2x+5} \, dx \) using the substitution \( u = 2x + 5 \), we first find the differential \( du \). We differentiate \( u \) with respect to \( x \): \[ du = 2 \, dx \quad \Rightarrow \quad dx = \frac{du}{2} \] Now, we can rewrite the integral in terms of \( u \): \[ \int \sqrt{2x + 5} \, dx = \int \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/2} \, du \] Next, we integrate \( u^{1/2} \): \[ \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{3} u^{3/2} \] Finally, substituting back \( u = 2x + 5 \): \[ \int \sqrt{2x + 5} \, dx = \frac{1}{3} (2x + 5)^{3/2} + C \] So, we have: \[ \int \sqrt{2x + 5} \, dx = \frac{1}{3} (2x + 5)^{3/2} + C \]